Answer: The position vectors are

\(\mathbf{p}=\mathbf{6i}+\mathbf{2j}+\mathbf{7k}+\frac{5}{2}(\mathbf{i}+\mathbf{j})=\frac{17}{2}\mathbf{i}+\frac{9}{2}\mathbf{j}+7\mathbf{k}\)

and

\(\mathbf{q}=\mathbf{4i}+\mathbf{4j}+\frac{5}{2}(-6\mathbf{j}+\mathbf{k})=4\mathbf{i}-11\mathbf{j}+\frac{5}{2}\mathbf{k}\).

Let

\(P=(6+\lambda,\,2+\lambda,\,7)\) on \(l_1\), since its direction vector is \((1,1,0)\),

and \(Q=(4,\,4-6\mu,\,\mu)\) on \(l_2\), since its direction vector is \((0,-6,1)\).

Then

\(\overrightarrow{PQ}=Q-P=( -2-\lambda,\,2-6\mu-\lambda,\,\mu-7)\).

Because \(PQ\) is perpendicular to both lines, it is perpendicular to both direction vectors.

So

\(\overrightarrow{PQ}\cdot(1,1,0)=0\), giving

\((-2-\lambda)+(2-6\mu-\lambda)=0\)

\(\Rightarrow -2\lambda-6\mu=0\)

\(\Rightarrow \lambda+3\mu=0\).

Also

\(\overrightarrow{PQ}\cdot(0,-6,1)=0\), so

\(-6(2-6\mu-\lambda)+(\mu-7)=0\)

\(\Rightarrow -12+36\mu+6\lambda+\mu-7=0\)

\(\Rightarrow 6\lambda+37\mu=19\).

Using \(\lambda=-3\mu\) in \(6\lambda+37\mu=19\):

\(-18\mu+37\mu=19\)

\(\Rightarrow 19\mu=19\)

\(\Rightarrow \mu=1\), hence \(\lambda=-3\).

Therefore

\(P=(6-3,\,2-3,\,7)=(3,-1,7)\),

and

\(Q=(4,\,4-6,\,1)=(4,-2,1)\).

So the required position vectors are

\(\boxed{\mathbf{p}=3\mathbf{i}-\mathbf{j}+7\mathbf{k}}\) and \(\boxed{\mathbf{q}=4\mathbf{i}-2\mathbf{j}+\mathbf{k}}\).